R help - Jan 2009 - Linked count between matrix columns

Hello,

I create this array:

x <- cbind(c(1:4, rep(0,10)), c(rep(0,4), 1:2, rep(3,6), 4,5))

       [,1] [,2]
  [1,]    1    0
  [2,]    2    0
  [3,]    3    0
  [4,]    4    0
  [5,]    0    1
  [6,]    0    2
  [7,]    0    3
  [8,]    0    3
  [9,]    0    3
[10,]    0    3
[11,]    0    3
[12,]    0    3
[13,]    0    4
[14,]    0    5

I would like to do the following in vector syntax:

for rows where the first column is not 0, put into the second column  
the number of times the value of the first column appears in the  
second column of rows where the value in the first row is 0

I'm not sure this sounds super clear, so I will show what I want to get:

       [,1] [,2]
  [1,]    1    1
  [2,]    2    1
  [3,]    3    6
  [4,]    4    1
  [5,]    0    1
  [6,]    0    2
  [7,]    0    3
  [8,]    0    3
  [9,]    0    3
[10,]    0    3
[11,]    0    3
[12,]    0    3
[13,]    0    4
[14,]    0    5

So for example, x[3,2] = 6, because length(x[x[,1]==0 & x[,2]==3,2]) = 6

I have tried this:

x[x[,1]!=0,2] <- length(x[x[,1]==0 & x[,2] %in% which(x[,1]!=0),2])

but it does not work correctly as it put the same value in the changed  
rows.

Thanks for your help!

Guillaume

Try this:

x[,2][x[,1][x[,1] > 0]] <- table(x[,2])[as.character(x[,1][x[,1] > 0])]

On Thu, Jan 15, 2009 at 10:36 AM, Guillaume Chapron <
carnivorescience@gmail.com> wrote:
> Hello,
>
> I create this array:
>
> x <- cbind(c(1:4, rep(0,10)), c(rep(0,4), 1:2, rep(3,6), 4,5))
>
>      [,1] [,2]
>  [1,]    1    0
>  [2,]    2    0
>  [3,]    3    0
>  [4,]    4    0
>  [5,]    0    1
>  [6,]    0    2
>  [7,]    0    3
>  [8,]    0    3
>  [9,]    0    3
> [10,]    0    3
> [11,]    0    3
> [12,]    0    3
> [13,]    0    4
> [14,]    0    5
>
> I would like to do the following in vector syntax:
>
> for rows where the first column is not 0, put into the second column the
> number of times the value of the first column appears in the second column
> of rows where the value in the first row is 0
>
> I'm not sure this sounds super clear, so I will show what I want to
get:
>
>      [,1] [,2]
>  [1,]    1    1
>  [2,]    2    1
>  [3,]    3    6
>  [4,]    4    1
>  [5,]    0    1
>  [6,]    0    2
>  [7,]    0    3
>  [8,]    0    3
>  [9,]    0    3
> [10,]    0    3
> [11,]    0    3
> [12,]    0    3
> [13,]    0    4
> [14,]    0    5
>
> So for example, x[3,2] = 6, because length(x[x[,1]==0 & x[,2]==3,2]) =
6
>
> I have tried this:
>
> x[x[,1]!=0,2] <- length(x[x[,1]==0 & x[,2] %in% which(x[,1]!=0),2])
>
> but it does not work correctly as it put the same value in the changed
> rows.
>
> Thanks for your help!
>
> Guillaume
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>


-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O

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Thank you! This is exactly what I wanted. Could you please explain the  
logic behind your code?

> x[,2][x[,1][x[,1] > 0]] <- table(x[,2])[as.character(x[,1][x[,1] >
> 0])]